If tan 5 2 and cos 0, then sin

tan(5 + =tan( )=2, and cos >0 in the first quadrant, sin(- =-sin( )=-root(1-cos2)-root[1-1(1+tan2( )=-root[1-1(1+4)]-2 5*root5

sin(-π=-sin(π-=-sinθ

tan(5 + =tan( +=tan =2sec 2 =1+tan 2 =1+4=5 because cos >0

sec = root number 5

cosθ=1/5

sin = root number 1-cos 2 = 2 root number 6 5 so sin (- = -2 root number 6 5

tan(5 + =tan =2,1 cos) 2=(sec ) 2=1+(tan ) 2=5,cos ) 2=1 5,sin ) 2=4 5, and because cos >0 and tan =2>0, so sin >0, so sin(- =-sin =-2 (root number 5)

tan(5 + =tan( +=tan =2 Description in one and three quadrants.

And because cos >0 is in the first quadrant.

So. sin(- =-sin( -=-sin Countable sin 2=4 5

It is known whether it is: tan( +=2tan

If yes, then:

3sinβ=sin(2α+β

=>3sin(α+=sin(α+==>3sin(α+cosα-3cos(α+sinα=sin(α+cosα+cos(α+sinα

=>2sin( +=4cos( +sin <==>tan( +=2tan , which is consistent with the known reversibility of the above steps, so the proposition is true.

Please look at this, it will help you.

tan(α+=2tanα,sin(α+/cos(α+=2sin(α)/cos(α)sin(α+cos(α)=2cos(α+sin(α)2sin(α+cos(α)=4cos(α+sin(α)3sin(α+cos(α)3cos(α+sin(α)=sin(α+cos(α)cos(α+sin(α)

3sin[(α=sin(α+cos(α)cos(α+sin(α)

3sin( )=sin[( i.e.: 3sin =sin(2+.)

sin(α+2β)=sin[(α=sin(α+cosβ+cos(α+sinα

sinα=sin[(α=sin(α+cosβ-cos(α+sinβ

Substituting sin(+2)=3sin

2sin( +cos -4cos( +sin =0 divided by cos cos( + simplify.

5sin =sin(2 + get:5sin( +sin( +sin( +sin( +cos -sin cos( +sin( +cos +sin cos( +with the cover Yu Weizai cos( +de:5tan( +cos -sin =tan( +cos +sin put tan( +9 4 substitution, get the mountain roll:

45/4)cosα-sinα=(9/4)cosα+sinαtanα=9/2

a∈(-2,0)

a Troublemaker (0,3 2).

sin( -a)=-2 3 So the liquid disturbance is -a ( 3 Li Yun 2)cos( -a).